On Voevodsky's paper 'Cancellation theorem', Lemma 4.8, he stated that the map $$\begin{array}{ccc}\mathbb{G}_m\times\mathbb{G}_m&\longrightarrow&\mathbb{G}_m\times\mathbb{G}_m\\(x,y)&\longmapsto&(y,x^{-1})\end{array}$$ is $\mathbb{A}^1$-homotopic to the identity, i.e., one could find an explicit homotopy (finite correspondence) $$h:\mathbb{G}_m\times\mathbb{G}_m\times\mathbb{A}^1\longrightarrow\mathbb{G}_m\times\mathbb{G}_m$$ such that $h(x,y,0)=(x,y), h(x,y,1)=(y,x^{-1})$.

But I don't see how to find such a map.

Alternatively, he wants to show that the map $$\begin{array}{ccc}\mathbb{G}_m\times\mathbb{G}_m&\longrightarrow&\mathbb{G}_m\times\mathbb{G}_m\\(x,y)&\longmapsto&(y,x)\end{array}$$ is $\mathbb{A}^1$-homotopy to $$\begin{array}{ccc}\mathbb{G}_m\times\mathbb{G}_m&\longrightarrow&\mathbb{G}_m\times\mathbb{G}_m\\(x,y)&\longmapsto&(x, y^{-1})\end{array}$$.

On Voevodsky's paper 'Cancellation theorem', Lemma 4.8, he stated in the proof that the map $$\begin{array}{ccc}\mathbb{G}_m\times\mathbb{G}_m&\longrightarrow&\mathbb{G}_m\times\mathbb{G}_m\\(x,y)&\longmapsto&(y,x^{-1})\end{array}$$ is $\mathbb{A}^1$-homotopic to the identity, i.e., one could find an explicit homotopy (finite correspondence) $$h:\mathbb{G}_m\times\mathbb{G}_m\times\mathbb{A}^1\longrightarrow\mathbb{G}_m\times\mathbb{G}_m$$ such that $h(x,y,0)=(x,y), h(x,y,1)=(y,x^{-1})$.

But I don't see how to find such a map.

Alternatively, he wants to show that the map $$\begin{array}{ccc}\mathbb{G}_m\times\mathbb{G}_m&\longrightarrow&\mathbb{G}_m\times\mathbb{G}_m\\(x,y)&\longmapsto&(y,x)\end{array}$$ is $\mathbb{A}^1$-homotopy to $$\begin{array}{ccc}\mathbb{G}_m\times\mathbb{G}_m&\longrightarrow&\mathbb{G}_m\times\mathbb{G}_m\\(x,y)&\longmapsto&(x, y^{-1})\end{array}$$.

On Voevodsky's paper 'Cancellation theorem', Lemma 4.8, he stated that the map $$\begin{array}{ccc}\mathbb{G}_m\times\mathbb{G}_m&\longrightarrow&\mathbb{G}_m\times\mathbb{G}_m\\(x,y)&\longmapsto&(y,x^{-1})\end{array}$$ is $\mathbb{A}^1$-homotopic to the identity, i.e., one could find an explicit homotopy (finite correspondence) $$h:\mathbb{G}_m\times\mathbb{G}_m\times\mathbb{A}^1\longrightarrow\mathbb{G}_m\times\mathbb{G}_m$$ such that $h(x,y,0)=(x,y), h(x,y,1)=(y,x^{-1})$.

But I don't see how to find such a map.

On Voevodsky's paper 'Cancellation theorem', Lemma 4.8, he stated that the map $$\begin{array}{ccc}\mathbb{G}_m\times\mathbb{G}_m&\longrightarrow&\mathbb{G}_m\times\mathbb{G}_m\\(x,y)&\longmapsto&(y,x^{-1})\end{array}$$ is $\mathbb{A}^1$-homotopic to the identity, i.e., one could find an explicit homotopy (finite correspondence) $$h:\mathbb{G}_m\times\mathbb{G}_m\times\mathbb{A}^1\longrightarrow\mathbb{G}_m\times\mathbb{G}_m$$ such that $h(x,y,0)=(x,y), h(x,y,1)=(y,x^{-1})$.

But I don't see how to find such a map.

Alternatively, he wants to show that the map $$\begin{array}{ccc}\mathbb{G}_m\times\mathbb{G}_m&\longrightarrow&\mathbb{G}_m\times\mathbb{G}_m\\(x,y)&\longmapsto&(y,x)\end{array}$$ is $\mathbb{A}^1$-homotopy to $$\begin{array}{ccc}\mathbb{G}_m\times\mathbb{G}_m&\longrightarrow&\mathbb{G}_m\times\mathbb{G}_m\\(x,y)&\longmapsto&(x, y^{-1})\end{array}$$.